Discussion in 'Homework Help' started by ac_dc_1, Jan 27, 2013.

1. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
Hello! I am trying to answer the following question:

Design an circuit with operational amplifiers that has the following output function:

Vo=2*V1+10*V2+11*V3-V4

Shoud i use the inverting adder?

Using three different voltage sorces V1,V2,V3 and V4 ,an OpAmp and a 4 resistors(R1 to R4) and a feedbak resistor.

So the expression for the output of and adding inverter is given by:

V0=-(Rf/R1)-(Rf/R2)-(Rf/R3)-(Rf/R4)

So i could use Rf=10k ,R1=5k,R2=1k,R3=1,1k and R4=10k

The problem is then Vo=-2V1-10V2-11V3-V4....Shoud i use and non-inverting adder,conecing the resistors to the non-inverting input?

Thanks

2. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
You could do it with one op amp, but the math is messy.
With 2 op amps, it becomes simple. Your problem definition doesn't restrict you a single op amp.

3. ### WBahn Moderator

Mar 31, 2012
24,338
7,600
Are these the constraints you have to work with? If so, then are V1, V2 and V3 the "three different voltage sources" referred to? What is V4? Some reference voltage?

4. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
V1,V2,V3 and V4 are reference voltages.Using two opamps,i have to do an inverting adder with V1,V2 and V3?
The problem is that in the function that i am looking for all terms are positive except for V4...So using an inverting adder i will get -2V1-10V3-11V3....How can i make the terms positive?Is there another configuration of a non inverting adder?

Thanks

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
This is homework help, so I'm not going to tell you the solution, but here is a big hint.
Use two cascaded inverting summing amplifiers. Keep in mind that you don't have to do all the summing in the first amplifier.

6. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
I think you could use one opamp to create 11*(V3-V4) and another to do the other summing.
The problem is that in the function that i am looking for all terms are positive except for V4...So using an inverting adder i will get -2V1-10V3-11V3....How can i make the terms positive?Is there another configuration of a non inverting adder?

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Here's a bigger hint:

Vout = -(-2V1-10V2-11V3)-V4

8. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0

So i use one inverting adder with three resistors (i can make Rf=10KR1=5k,R2=1k,R3=1,1k) referenced to V1,V2 and V3 to do -2V1-10V2-11V3

Then the output of the that inverting adder enters the inverting input of the second inverting adder where i have one resistor (R4=10k) referenced to V4.

Is that it?

Thanks

9. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
You have to have a resistor between the output of the 1st op amp to the -input of the 2nd op amp, and a feedback resistor.
What are their values?

Also, R3=1.1k Will not give you the correct answer. You need to think a little harder and do the math.

10. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0

The feedback resistor of both amplifiers will be equal to 10k(value that i defined)

The value of R3 was incorrect as you mentioned it should be something like 0.9k

About the resistor that conects the output of the first op amp to the second i have no idea how to calculate its value..I thought i could conect de Vo of the 1st Op Amp to the inverting input of the second.What are the conditions for calculating that resistor?

Thanks

11. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
You understand the concept of summing in the 1st amplifier. You are summing two terms in the second amplifier, right? What are those terms? Think, man. I believe you are leaning too heavily on me, and not using your head. I won't be here when you get in a workplace environment.

12. ### WBahn Moderator

Mar 31, 2012
24,338
7,600
What is the mathematical relationship for the output of the first amp, including all the scaling? Call the V5. You second circuit is just another inverting adder that is using V5 and V4. What is the output of the second circuit in terms of V4 and V5, including all the scaling due to the resistors in that circuit? Now substitute in the first expression (what V5 is in terms of V1, V2 and V3) and you now can see how each term is affected by which resistors.

13. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
In the second amplifier i am summing the output of the first(composed by the sum of V1,V2 and V3) with V4

Lets see my function is:

Vout = -(-2V1-10V2-11V3)-V4

From the expression i conclude that there is not any linear relation between the output of the first OpAmp and the second... Vout = -(-2V1-10V2-11V3)-V4

Quoting this:"What is the output of the second circuit in terms of V4 and V5, including all the scaling due to the resistors in that circuit?"

I think that what you are saying is that i have somehing like Vo=V4-V5 so therefore i should have a resitor of 1k conecting the first Op Amp and the second.Is that it?

Thanks

14. ### WBahn Moderator

Mar 31, 2012
24,338
7,600
The value of the resistor connecting them depends on the scaling in the first opamp circuit and also the value of the feedback resistor in the second.

The output of the second opamp is a linear combination of the output from the first opamp and V4.

15. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
You are sloppy with your signs. You don't have "something like Vo=V4-V5". Draw a schematic for us, with both op amps and all resistor values that you know, and with the equations for the outputs of both op amps shown. Post the schematic. If that doesn't give you an "aha!" moment, tell us where you are stuck.

16. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
Yes i was a bit sloppy with my signs.I would have -(-V4-V5)=V4-V5,which is the same thing..Note that i was saying that V4=-2V1+10V2+11V3 and V5=-V4...
Anyway, i draw (sorry for the quality of the draw) and unfortunately i did not have a "aha!" moment because i have never seen any montage like this or speak about it in classes.

I did write the equations because its easy to do it here so:

Vo1(output of the first OpAmp)=-V1*(Rf/R1)-V2*(Rf/R2)-V3*(Rf/R3)
Vo2(output of the second OpAmp)=-Vo1*(Rf/Rconection)-V4*(Rf/R4)

So Vo=-(Vo1-Vo2)=-Vo1+Vo2=-(-2V1-10V2-11V3)-V4=(2V1+10V2+11V3)-V4

For the values of the resistors i would still say:

Rf=10k

R1=5K
R2=1K
R3=0.9K
R4=10K

Rconection(do not konw but..by what seems logic to me not to change the relationships)=1k

Thanks

File size:
59.3 KB
Views:
51
17. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
I don't like giving answers to problems, but you seem to be stuck. You must have a mental block on this problem.
Tell us if you understand the attached schematic. If not, tell us what parts are giving you problems.

• ###### ac_dc_1's problem.png
File size:
35.1 KB
Views:
48
ac_dc_1 likes this.
18. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
Ok i got it now....The relationship between the conection resistor and the feedback resistor of the second OpAmp has to be equal to 1,to as i said no to change the function..

The problem was that i was chosing the same values for RF1 and RF2.....But if i used a resistor of 1k to conect the two OpAmps and a feeback resistor of 1k in the second OpAmp it would also be correct, rigth?

Thanks

Last edited: Jan 29, 2013
19. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Yes, you could use R4=R5=Rf2=1k.